1. B
Just move the triangle to the right 2 units and down 3 units and the new coordinates are the answer
2. C
There are 2 cups of flour for every batch
Answer:
positive
Step-by-step explanation:
if x is positive then for all values 5x^3 which is always>-2x will be positive and therefore answer always positive
if x is negative then 5x^3 will always be positive and -2(X) will cancel the two minuses -- and will add therefore positive
Answer:

Step-by-step explanation:
we know that
The directrix of the parabola is perpendicular to the axis of symmetry of the parabola
In this problem the directrix is y=-3.5
so
The axis of symmetry is parallel to the y-axis
we have a vertical parabola
Also, the vertex is at the origin
That means-----> the parabola open upward
The equation of a vertical parabola can be written as

where
p is the distance between the vertex and the directrix
In this problem
the distance between the vertex and the directrix is 3.5
----> the value of p is positive because the parabola open upward
substitute


isolate the variable y

Answer:
((4xysqrt(xy))-y)/(x-2x^2sqrt(xy))
Step-by-step explanation:
Treat y as a function of x and use the chain rule.
Use the chain rule and product rule:
(y+xy')/2sqrt(xy) = d/dx[8+x^2y]
(y+xy')/2sqrt(xy) = d/dx[x^2y]
Use product rule
(y+xy')/2sqrt(xy) = 2xy + x^2y'
Now solve for y'
y+xy' = (2xy)(2sqrt(xy))+ (x^2y')(2sqrt(xy))
xy'-2x^2y'sqrt(xy) = (2xy)(2sqrt(xy))-y
y'(x-2x^2sqrt(xy))
y' = ((4xysqrt(xy))-y)/(x-2x^2sqrt(xy))